162 ANALYTICAL MECHANICS 12. Find the moment of inertia of an ellipsoid with respect to, (a) one of its axes, (b) a tangent at one end of one of the axes parallel to the other. 13. In the preceding problem suppose the body to be an ellipsoidal shell of negligible thickness. 125. General Method. — The special methods which have been discussed in the last few pages are desirable but not necessary for finding the moments of inertia of bodies. Instead of selecting special types of elements of mass for each type of bodies and then making use of the various theorems we can use the general expressions for dm which were given on p. 147 and obtain the moment of inertia directly from equation (II). As an illustration of this general method consider the moment of inertia of a sphere with respect to a diameter. It is evident from the symmetry of the body that the moment of inertia about a diameter equals eight times that of an octant about one of its straight edges. (a) Let the octant be taken as shown in Fig. 81, and be referred to Cartesian coordinates, then the equations of the bounding surfaces are & + y2 + z2 = a2, x = 0, y = 0, and z = 0. Hence taking the z-axis as the axis of reference we have nffl I = C r*dm JQ /** pV^tf fV^-xt-yi 888 8rJo Jo Jo <y* + **)dxdydz = | ma2. (b) Let the octant be referred to spherical coordinates, Fig. 80, then the equations of the bounding surfaces are r = a, B = , <£ = 0, $ = , and p = a. Therefore / = • = f (P2 ~ P2 sin2 6 cos2 <t>} dm J 0 TT 7T = 8r f2 f2 f V (sin 0 _ Sin3 0 COS2^) M d(f) dp «/0 ** 0 •/O